Angle Between Two Lines

Acute angle θ between two lines with slopes m₁ and m₂. The obtuse angle is π − θ.

Let the two lines have inclinations $\theta_1$ and $\theta_2$, so $m_1 = \tan\theta_1$ and $m_2 = \tan\theta_2$. The angle from line 1 to line 2 is $\phi = \theta_2 - \theta_1$. By the tangent subtraction identity:

$$\tan\phi = \frac{\tan\theta_2 - \tan\theta_1}{1 + \tan\theta_1\tan\theta_2} = \frac{m_2 - m_1}{1 + m_1 m_2}$$

Two intersecting lines form two pairs of supplementary angles. Taking the acute angle $\theta$:

$$\boxed{\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|}$$

Special cases

Parallel: $\tan\theta = 0 \Rightarrow m_1 = m_2$.

Perpendicular: $\tan\theta \to \infty \Rightarrow 1 + m_1m_2 = 0 \Rightarrow m_1m_2 = -1$.

General form version

For $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, substitute $m_i = -a_i/b_i$:

$$\tan\theta = \left|\frac{a_1b_2 - a_2b_1}{a_1a_2 + b_1b_2}\right|$$ > > $$ > \cos\theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{|\vec{d_1}||\vec{d_2}|} > $$ > > The absolute value gives the acute angle. With $\vec{d_1} = (1, m_1)$ and $\vec{d_2} = (1, m_2)$: > > $$ > \cos\theta = \frac{|1 + m_1 m_2|}{\sqrt{1+m_1^2}\sqrt{1+m_2^2}} > $$ > > The $\tan\theta$ formula and this $\cos\theta$ formula are equivalent — related by the identity $\tan\theta = \sin\theta/\cos\theta$ with $\sin\theta = |\vec{d_1} \times \vec{d_2}|/(|\vec{d_1}||\vec{d_2}|)$. In 3D, the dot product form generalises directly; the $\tan\theta$ slope form does not.